# Proof for sup and inf of $A=\{1-1/n:n\in\mathbb{N}\}⊂\mathbb{Q}$

Let $A=\{1-1/n:n\in\mathbb{N}\}⊂\mathbb{Q}$

I want to proof the supremum and infimum of the set $A$ is. I can intuitively see that it is $sup(A)=1$ and $inf(A)=0$.

I know I need to proof that 1 is in the upper bound and 0 is in the lower bound, and then that it is the smallest upper and largest lower bound.

The upper bound of $A$ is $1$, as $1/n$ tends to 0 when $n$ gets larger. And the lower bound is $0$ for the same reason.

But how do I reason/proof that they are the smallest upper and largest lower bound?

• Suppose that there is a smaller upper bound (resp. larger lower bound) and arrive at a contradiction. – wjm Oct 2 '17 at 19:04
• Can you use the result that $a$ is the supremum of $A$ if and only if $a$ is an upper bound for $A$ and for all $\varepsilon>0$ we have $(a-\varepsilon,a]\cap A\neq\emptyset$? (similar result for $\inf(A)$). – Dave Oct 2 '17 at 19:23

You want to show that for all $x<1$, $x$ is not an upper bound of $A$, which means there exists some $a\in A$ such that $a>x$. Finding this $a$ will have to depend on $x$.
Similarly, you want to show that for all $x>0$, $x$ is not a lower bound of $A$, meaning there exists some $a\in A$ such that $a<x$. Finding this $a$ will be easy.
• So if I let $a=1$ I have $1>x$. Is writing this (in a more proper way) a correct way of showing that $sup(A)=1$? – Sirmimer Oct 2 '17 at 19:28
• @Sirmimer Well, you can't take $a=1$, since $1\not\in A$. – Chris Culter Oct 2 '17 at 19:32
• I thought $A=[0,1]$ as $1/n$ approaches 0 and therefore you have $1-0=1$? – Sirmimer Oct 2 '17 at 19:36
• @Sirmimer $1/n$ approaches $0$, but there is no value of $n$ for which it equals $0$. So $1$ is not a member of the set $A$. – Chris Culter Oct 2 '17 at 19:56